Integral Calculus 201-SN3-RE • CEGEP, Montreal

Revision sheet: antiderivatives and the indefinite integral (201-SN3-RE)

Q: Why do you have to add a constant to an antiderivative?

Because differentiating wipes out constants: all the functions that differ by a constant have the same derivative. The indefinite integral therefore means the whole set of these functions, not one of them. Without the constant, you cannot answer the questions that impose a value at a point.

Q: Is there a product rule for integrals?

No, and it is the most common mistake in the chapter. The integral of a product is not the product of the integrals. The technique that comes closest is integration by parts, which comes from the derivative of a product read backwards, but it does not distribute and cannot be applied mechanically.

Q: How do you know if a substitution will work?

Choose an inner block, compute its derivative, then look for that derivative in the integrand. If it appears as a factor, even up to a constant, the substitution goes through. If the gap is a variable and not a constant, you have to change technique.

Q: Why do we write an absolute value in the antiderivative of one over x?

Because the reciprocal function is also defined for negative values, whereas the natural logarithm is not. The absolute value extends the antiderivative to both intervals. Remember too that the constant can differ from one interval to the other, since the two branches are separated by an asymptote.

Q: Do all functions have an antiderivative you can compute?

Every continuous function has antiderivatives, but some cannot be expressed with the usual functions. That is the case for the exponential of minus x squared, for sine divided by x, and for one over the logarithm of x. Recognizing them keeps you from wasting ten minutes of an exam looking for a formula that does not exist.

This sheet does not stop at the table of antiderivatives, which takes a minute to learn and is not where marks are lost. It deals with what decides the grade on the first 201-SN3-RE exam, where almost every mistake is a rule invented by analogy with differentiation, and where the check by differentiation, which costs nothing, is almost never done.

It is written for CEGEP students in the Science program (Sciences de la nature) taking Integral Calculus, 201-SN3-RE (formerly 201-NYB). The corrected exercise set on the same chapter then puts every reflex to the test.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

An antiderivative is CHECKED by differentiating it: this is the only chapter of the course where you can verify every answer yourself in ten seconds. Everything that costs marks here comes from inventing a rule that does not exist, or from forgetting that the answer is a FAMILY of functions, not a single function.

This chapter is part of 201-SN3-RE, Integral Calculus

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Antiderivative of a function
Indefinite integral
Constant of integration
Linearity of the integral
Table of antiderivatives
Direct substitution
Change of variable
Natural logarithm as an antiderivative
Method of undetermined coefficients

The essentials

One antiderivative, one family, one constant

• $F$ is an antiderivative of $f$ if $F'=f$ . If $F$ is one, so is $F+C$ , for every real $C$ .
• $\int f(x)\,dx=F(x)+C$ : the result is a FAMILY of functions, never a single one.
• Two antiderivatives of the same function differ by a constant ON AN INTERVAL, and this is a consequence of the mean value theorem.
• Geometrically: all the curves in the family are obtained from one another by VERTICAL translation, and therefore have the same slope at each $x$ -value.

Three antiderivatives of

f(x)=x

for

C=-3

0

and

3

: the curves are vertical translates and their tangent lines at the same

x

-value are PARALLEL, since they have the same derivative.

The $+C$ is not decoration: it is worth half a mark on every indefinite integral, and an initial-condition question is impossible without it.

The only two rules, and the ones that do not exist

• Linearity: $\int\left(af+bg\right)=a\int f+b\int g$ . That is all that distributes.
• $\int x^{n}\,dx=\frac{x^{n+1}}{n+1}+C$ for $n\neq-1$ , and $\int\frac{dx}{x}=\ln\left|x\right|+C$ for the case $n=-1$ .
• There is NO product rule: $\int fg\neq\left(\int f\right)\left(\int g\right)$ .
• There is NO quotient rule: $\int\frac{f}{g}\neq\frac{\int f}{\int g}$ .
• There is NO automatic reverse chain rule: $\int f(g(x))\,dx$ calls for a substitution, which only works if $g'$ is present.

A counterexample to remember: $\int x\cdot x\,dx=\frac{x^{3}}{3}$ , whereas $\left(\int x\,dx\right)^{2}=\frac{x^{4}}{4}$ .

Composite forms, the ones you read at a glance

• $\int u'u^{n}\,dx=\frac{u^{n+1}}{n+1}+C$ , $\int\frac{u'}{u}\,dx=\ln\left|u\right|+C$ , $\int u'e^{u}\,dx=e^{u}+C$ .
• $\int u'\cos u\,dx=\sin u+C$ , $\int u'\sin u\,dx=-\cos u+C$ , $\int\frac{u'}{1+u^{2}}\,dx=\arctan u+C$ .
• They all say the same thing: the inner derivative MUST be present as a factor, up to a multiplicative constant.
• A missing constant can be made up by multiplying and dividing; a missing variable can NEVER be made up.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

The common antiderivatives

The table is read from left to right, and CHECKED from right to left: differentiating the right-hand column must give back the left-hand one. This is the only chapter of the course where the answer can be checked in ten seconds.

$f(x)$	Where it holds	$\int f(x)\,dx$
$x^{n}$ , $n\neq-1$	$n$ any real number	$\frac{x^{n+1}}{n+1}+C$
Example: $\int x^{3}\,dx=\frac{x^{4}}{4}+C$ , so $\int_{0}^{2}x^{3}\,dx=\frac{16}{4}=4$ .
$\frac{1}{x}$	$x\neq 0$	$\ln\left\|x\right\|+C$
Example: This is the case $n=-1$ , the one the previous row does not cover. $\int_{1}^{\mathrm{e}}\frac{dx}{x}=\ln \mathrm{e}-\ln 1=1$ .
$\mathrm{e}^{x}$	everywhere	$\mathrm{e}^{x}+C$
Example: $\int_{0}^{1}\mathrm{e}^{x}\,dx=\mathrm{e}-1\approx 1.718$ .
$a^{x}$ , $a>0$	everywhere	$\frac{a^{x}}{\ln a}+C$
Example: $\int_{0}^{1}2^{x}\,dx=\frac{2-1}{\ln 2}=\frac{1}{\ln 2}\approx 1.443$ . The $\ln a$ is in the DENOMINATOR here, in the numerator in the derivative.
$\cos x$	in radians	$\sin x+C$
Example: $\int_{0}^{\pi/2}\cos x\,dx=\sin\frac{\pi}{2}-\sin 0=1$ , the area under an arch of cosine up to its first zero.
$\sin x$	in radians	$-\cos x+C$
Example: The minus sign moves from the derivative to the antiderivative: $\int_{0}^{\pi}\sin x\,dx=-\cos\pi+\cos 0=2$ .
$\sec^{2}x$	$x\neq\frac{\pi}{2}+k\pi$	$\tan x+C$
Example: $\int_{0}^{\pi/4}\sec^{2}x\,dx=\tan\frac{\pi}{4}=1$ .
$\frac{1}{1+x^{2}}$	everywhere	$\arctan x+C$
Example: $\int_{0}^{1}\frac{dx}{1+x^{2}}=\arctan 1=\frac{\pi}{4}\approx 0.785$ . No substitution gives this row, it has to be recognized.
$\frac{1}{\sqrt{1-x^{2}}}$	$\left\|x\right\|<1$	$\arcsin x+C$
Example: $\int_{0}^{1/2}\frac{dx}{\sqrt{1-x^{2}}}=\arcsin\frac{1}{2}=\frac{\pi}{6}\approx 0.524$ .

The $+C$ is not decoration: without it, the answer is ONE function where the question asks for a family, and the half mark is lost on every question.

Composite forms: the $u'$ must be there

The six rows all say the same thing, and it is the only thing to know about substitution: the inner derivative must be present AS A FACTOR, up to a multiplicative constant.

Integrand	What you set	Antiderivative
$u'u^{n}$ , $n\neq-1$	$u$ and its $u'$	$\frac{u^{n+1}}{n+1}+C$
Example: $\int 2x\left(x^{2}+1\right)^{3}dx=\frac{\left(x^{2}+1\right)^{4}}{4}+C$ , and from $0$ to $1$ this equals $\frac{16-1}{4}=3.75$ .
$\frac{u'}{u}$	$u$ in the denominator	$\ln\left\|u\right\|+C$
Example: $\int_{0}^{1}\frac{2x}{x^{2}+1}dx=\ln 2-\ln 1=\ln 2\approx 0.693$ .
$u'\mathrm{e}^{u}$	$u$ in the exponent	$\mathrm{e}^{u}+C$
Example: $\int_{0}^{1}2x\,\mathrm{e}^{x^{2}}dx=\mathrm{e}^{1}-\mathrm{e}^{0}=\mathrm{e}-1\approx 1.718$ .
$u'\cos u$	$u$ inside the cosine	$\sin u+C$
Example: $\int_{0}^{\pi/4}2\cos(2x)\,dx=\sin\frac{\pi}{2}-\sin 0=1$ : here $u=2x$ and $u'=2$ , present as is.
$\frac{u'}{1+u^{2}}$	$u$ squared	$\arctan u+C$
Example: $\int_{0}^{1/2}\frac{2\,dx}{1+4x^{2}}=\arctan 1=\frac{\pi}{4}\approx 0.785$ , with $u=2x$ .
A CONSTANT is missing	multiply and divide	the same antiderivative, divided by that constant
Example: $\int_{0}^{1}x\,\mathrm{e}^{x^{2}}dx=\frac{1}{2}\int_{0}^{1}2x\,\mathrm{e}^{x^{2}}dx=\frac{\mathrm{e}-1}{2}\approx 0.859$ .

A missing constant can be made up, a missing VARIABLE never: $\int \mathrm{e}^{x^{2}}dx$ does not reduce to any of these rows, and this integral in fact has no elementary antiderivative.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Forgetting the constant of integration

half a mark per integral, and the whole initial-condition question

What not to write

$\int 3x^{2}\,dx=x^{3}$

What to write

$\int 3x^{2}\,dx=x^{3}+C$

All the curves

x^{3}+C

have derivative

3x^{2}

; only one passes through

(1, 5)

, the one with

C=4

. Without the

+C

, the initial condition cannot be met.

Why: The indefinite integral means the set of all antiderivatives, not one of them. A question like “find $F$ such that $F(1)=5$ ” becomes impossible without the $C$ , and such questions often make up a third of the first exam.

2. Integrating a product factor by factor

the whole question

What not to write

$\int x\cos x\,dx=\frac{x^{2}}{2}\sin x$

What to write

$\int x\cos x\,dx=x\sin x+\cos x+C$ , obtained by integration by parts

Why: Immediate check: the derivative of $\frac{x^{2}}{2}\sin x$ is $x\sin x+\frac{x^{2}}{2}\cos x$ , which is not the integrand. The derivative of $x\sin x+\cos x$ is $\sin x+x\cos x-\sin x=x\cos x$ ✓. Thirty seconds of differentiation rule out the wrong answer.

3. Writing the antiderivative of one over x without an absolute value

1 mark, and a whole improper-integral question

What not to write

$\int\frac{dx}{x}=\ln x+C$

What to write

$\int\frac{dx}{x}=\ln\left|x\right|+C$ , with a DIFFERENT constant on each of the two intervals

\ln\left|x\right|

lives on TWO intervals separated by an asymptote: nothing links the two branches, so the constant of integration can be different on each.

Why: $\ln x$ only exists for $x>0$ , whereas $\frac{1}{x}$ is also defined for $x<0$ . More subtle: since the domain is cut in two, the constant can differ from one piece to the other, and an initial condition given on the left places no constraint at all on the right-hand branch.

4. Losing the $dx$ while substituting

the whole question, since the integral obtained mixes two variables

What not to write

“I set $u=x^{2}+1$ , so $\int x\sqrt{x^{2}+1}\,dx=\int x\sqrt{u}\,du$ .”

What to write

“ $u=x^{2}+1$ gives $du=2x\,dx$ , so $x\,dx=\frac{du}{2}$ and the integral becomes $\frac{1}{2}\int\sqrt{u}\,du$ .”

Why: An integral can contain only ONE variable. Leaving an $x$ next to a $du$ produces an expression that means nothing, and the marker stops reading there. The safe move: write $du$ in full, isolate what replaces the rest of the integrand, and check that no $x$ survives.

5. Forcing a substitution when the inner derivative is missing

ten minutes of exam time, and often the whole question

What not to write

“I set $u=x^{2}+1$ in $\int\sqrt{x^{2}+1}\,dx$ , so $dx=\frac{du}{2x}$ and I am done.”

What to write

“There is no $x$ as a factor: the substitution goes nowhere. This integral calls for a TRIGONOMETRIC substitution, $x=\tan\theta$ .”

Why: Dividing by $2x$ brings back an $x$ that nothing will remove: you go round in circles. The difference between $\int x\sqrt{x^{2}+1}\,dx$ , immediate, and $\int\sqrt{x^{2}+1}\,dx$ , hard, comes down to a single factor $x$ . Spotting it before you start saves exam time.

6. Not returning to the original variable

1 mark, every time

What not to write

$\int x\sqrt{x^{2}+1}\,dx=\frac{u^{3/2}}{3}+C$

What to write

$\int x\sqrt{x^{2}+1}\,dx=\frac{\left(x^{2}+1\right)^{3/2}}{3}+C$

Why: The variable $u$ is an intermediate tool, it does not exist in the question. An answer in $u$ is incomplete, like an equation solved for an auxiliary unknown that is never translated back. The reflex: the last line of a substitution is always the return to $x$ .

7. Believing that every elementary function has an elementary antiderivative

all the time spent searching

What not to write

“ $\int e^{-x^{2}}\,dx$ must be expressible with the usual functions, I am still looking.”

What to write

“ $e^{-x^{2}}$ is continuous, so it DOES have antiderivatives, but none of them can be expressed with the usual functions.”

e^{-x^{2}}

is continuous: its antiderivative equal to zero at

0

exists, it is increasing, and it is plotted here point by point. It simply has no elementary formula.

Why: Existence and computability are two different things: every continuous function has antiderivatives by the fundamental theorem, but $e^{-x^{2}}$ , $\frac{\sin x}{x}$ and $\frac{1}{\ln x}$ have no elementary one. Knowing this short list keeps you from persisting in vain on the exam.

8. Simplifying an antiderivative without looking at the domain

1 mark for rigour, more on a definite integral that crosses the point

What not to write

$\int\frac{x^{2}-4}{x-2}\,dx=\int(x+2)\,dx=\frac{x^{2}}{2}+2x+C$ on $\mathbb{R}$

What to write

The simplification only holds for $x\neq 2$ : the antiderivative is defined on $\left(-\infty, 2\right)$ and on $\left(2, +\infty\right)$ , with two possible constants.

The quotient agrees with

x+2

everywhere EXCEPT at

2

, where it is undefined: two intervals, so two possible constants.

Why: The integrand is not defined at $2$ , so neither is the antiderivative. Ignoring this detail later leads to computing $\int_{1}^{3}$ as if nothing were wrong, when the integral is improper and the result is worthless.

Which method to choose

Will a substitution work?

Spot an inner block $u$ , compute $u'$ , then look for $u'$ in the integrand. This test, and this test alone, decides.

If $u'$ is present as a factor, exactly → immediate substitution

Example: $\int 2x\,e^{x^{2}}\,dx$ with $u=x^{2}$ gives $e^{x^{2}}+C$
If $u'$ is present up to a CONSTANT → substitution after adjustment: multiply and divide by that constant

Example: $\int x\,e^{x^{2}}\,dx=\frac{1}{2}\int 2x\,e^{x^{2}}\,dx=\frac{e^{x^{2}}}{2}+C$
If the integrand has the form $\frac{u'}{u}$ → antiderivative $\ln\left|u\right|$ , to be recognized without any calculation

Example: $\int\frac{2x+3}{x^{2}+3x+7}\,dx=\ln\left|x^{2}+3x+7\right|+C$
If $u'$ is missing and the gap is a VARIABLE → simple substitution fails: move on to parts, partial fractions or a trigonometric substitution

Example: $\int\sqrt{x^{2}+1}\,dx$ calls for $x=\tan\theta$
If the integrand is a quotient of polynomials whose numerator has a degree greater than or equal to that of the denominator → do the DIVISION first, then integrate the quotient and the remainder separately

Example: $\frac{x^{2}}{x+1}=x-1+\frac{1}{x+1}$

When two substitutions seem possible, choose the one whose derivative shows up most clearly: it is almost always the contents of a root, of an exponential or of a denominator.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

A substitution, written up for the marker

When to use it: Any integral where an inner block and its derivative are both present.

1 Announce the change of variable: “Let $u=x^{2}+1$ .”
2 Write the full differential: “Then $du=2x\,dx$ , so $x\,dx=\frac{du}{2}$ .”
3 Rewrite the integral ENTIRELY in terms of $u$ , and check that no $x$ is left.
4 Integrate in $u$ , without forgetting the constant.
5 RETURN to the original variable, and conclude.

Concluding sentence

$\int x\sqrt{x^{2}+1}\,dx=\frac{1}{2}\int\sqrt{u}\,du=\frac{u^{3/2}}{3}+C=\frac{\left(x^{2}+1\right)^{3/2}}{3}+C$ .

The trap: Writing $du=2x$ instead of $du=2x\,dx$ . The differential carries the $dx$ ; without it, the replacement in the integral cannot be justified and the marker takes off the mark for rigour.

Marking: Usually half a mark for announcing the change, half a mark for the differential, one mark for the integral in $u$ , half a mark for the return to $x$ .

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Differentiate the answer, always

The antiderivative you found can be differentiated in thirty seconds, and its derivative must give back exactly the integrand. This is the only chapter of the course where checking is as fast as computing.

$\frac{\left(x^{2}+1\right)^{3/2}}{3}$ differentiates to $\frac{3}{2}\cdot\frac{\left(x^{2}+1\right)^{1/2}}{3}\cdot 2x=x\sqrt{x^{2}+1}$ ✓.
Check numerically at one point

Compute $\frac{F(a+0.001)-F(a-0.001)}{0.002}$ and compare it with $f(a)$ : the two must agree to three decimal places.

At $a=2$ : the numerical rate for $\frac{\left(x^{2}+1\right)^{3/2}}{3}$ gives $4.472$ , and $2\sqrt{5}=4.472$ ✓.
Check that no $x$ survives in an integral in $u$

After substituting, reread the integral: if an $x$ still appears, the substitution has not worked and you should change technique rather than carry on.

$\int x\sqrt{u}\,du$ is a meaningless expression: the next line has no chance of being right.
Check that the $+C$ is there

Reread every indefinite integral on your paper before handing it in and count the $+C$ 's: you need exactly as many as there are indefinite integrals.

Three indefinite integrals, three $+C$ 's. An initial-condition question without $+C$ has no solution at all.
Check an initial condition

When the question imposes $F(a)=b$ , substitute and solve for $C$ , then check that the final antiderivative does give $b$ at $a$ .

$F(x)=x^{3}+C$ with $F(1)=5$ gives $C=4$ , and $F(1)=1+4=5$ ✓.

The typical problem, taken apart

A substitution that resists, as on the first exam

Compute $\displaystyle\int\frac{x}{\sqrt{x+4}}\,dx$ .

The integrand

f(x)=\frac{x}{\sqrt{x+4}}

, defined for

x>-4

: the antiderivative found must have slope

f(0)=0

x=0

and slope

f(5)=\frac{5}{3}

x=5

, the two checks of the last step.

Step 1

Substitution test: with $u=x+4$ , we have $u'=1$ , present as a factor since $dx$ is. The substitution is therefore legitimate, even though the $x$ in the numerator is not $u'$ .

Why

The test is about the derivative of the INNER block, not about the overall look. Here $u'=1$ is always present, which makes the substitution possible as long as the lone $x$ is translated too.

Step 2

$u=x+4$ gives $du=dx$ and, above all, $x=u-4$ . The integral becomes $\int\frac{u-4}{\sqrt{u}}\,du$ .

Why

This is the move that unlocks it: express the REMAINING $x$ in terms of $u$ . A paper that does not think of inverting the substitution leaves an $x$ next to a $du$ and stops there.

Step 3

Split: $\frac{u-4}{\sqrt{u}}=\frac{u}{u^{1/2}}-\frac{4}{u^{1/2}}=u^{1/2}-4u^{-1/2}$ .

Why

Writing in powers makes the integral immediate. As long as the fraction is written with a root in the denominator, no formula from the table applies.

Step 4

$\int\left(u^{1/2}-4u^{-1/2}\right)du=\frac{2}{3}u^{3/2}-8u^{1/2}+C$ .

Why

Two applications of $\int u^{n}\,du=\frac{u^{n+1}}{n+1}$ , with $n=\frac{1}{2}$ then $n=-\frac{1}{2}$ . The second exponent puts $\frac{1}{2}$ in the denominator, hence the factor $2$ that turns into $8$ .

Step 5

Return to $x$ : $\frac{2}{3}(x+4)^{3/2}-8(x+4)^{1/2}+C$ , which can be factored as $\frac{2}{3}\sqrt{x+4}\left(x-8\right)+C$ .

Why

The factored form is not required, but it makes checking easier, and it is the one the answer key will show. The return to $x$ , on the other hand, is compulsory.

Step 6

Check by differentiation at $x=0$ : the antiderivative equals $\frac{2}{3}(2)(-8)=-\frac{32}{3}$ , and its numerical rate of change gives $0$ , which is indeed $\frac{0}{\sqrt{4}}$ ✓. At $x=5$ : the rate gives $1.667$ , and $\frac{5}{3}=1.667$ ✓.

Why

Two checkpoints are better than one: the first tests the value $0$ of the integrand, the second an arbitrary value. Together they catch a sign error as well as a forgotten factor.

The conclusion, written out

$\displaystyle\int\frac{x}{\sqrt{x+4}}\,dx=\frac{2}{3}(x+4)^{3/2}-8\sqrt{x+4}+C=\frac{2}{3}\sqrt{x+4}\,(x-8)+C$ .

The classic mistake on this problem: Giving up after $du=dx$ because an $x$ is left in the numerator. The substitution does not require the integrand to be entirely of the form $u'g(u)$ : it is enough to be able to rewrite everything in terms of $u$ , which $x=u-4$ makes possible here.

Learn by heart

• $\int f(x)\,dx=F(x)+C$ : the result is a FAMILY. The $+C$ is worth half a mark every time.
• Only LINEARITY distributes. No product rule, no quotient rule.
• $\int x^{n}\,dx=\frac{x^{n+1}}{n+1}+C$ except for $n=-1$ , where it is $\ln\left|x\right|+C$ .
• Composite forms require $u'$ AS A FACTOR: $\int u'u^{n}$ , $\int\frac{u'}{u}$ , $\int u'e^{u}$ .
• A missing constant can be made up, a missing variable never.
• Substitution: write $du$ with its $dx$ , rewrite everything in terms of $u$ , then RETURN to $x$ .
• Every continuous function has antiderivatives; $e^{-x^{2}}$ and $\frac{\sin x}{x}$ have no elementary one.
• Free check: differentiating the answer must give back the integrand.

Frequently asked questions

Why do you have to add a constant to an antiderivative?

Because differentiating wipes out constants: all the functions that differ by a constant have the same derivative. The indefinite integral therefore means the whole set of these functions, not one of them. Without the constant, you cannot answer the questions that impose a value at a point.

Is there a product rule for integrals?

No, and it is the most common mistake in the chapter. The integral of a product is not the product of the integrals. The technique that comes closest is integration by parts, which comes from the derivative of a product read backwards, but it does not distribute and cannot be applied mechanically.

How do you know if a substitution will work?

Choose an inner block, compute its derivative, then look for that derivative in the integrand. If it appears as a factor, even up to a constant, the substitution goes through. If the gap is a variable and not a constant, you have to change technique.

Why do we write an absolute value in the antiderivative of one over x?

Because the reciprocal function is also defined for negative values, whereas the natural logarithm is not. The absolute value extends the antiderivative to both intervals. Remember too that the constant can differ from one interval to the other, since the two branches are separated by an asymptote.

Do all functions have an antiderivative you can compute?

Every continuous function has antiderivatives, but some cannot be expressed with the usual functions. That is the case for the exponential of minus x squared, for sine divided by x, and for one over the logarithm of x. Recognizing them keeps you from wasting ten minutes of an exam looking for a formula that does not exist.

Practise it

Corrected exercises: Antiderivatives and the indefinite integral, 201-SN3-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

17 corrected exercises
170 points
225 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

Sending goes through your free space: that is how I can answer you and tell you when the exercise is added. An email is enough, no password. Create your space. Rather not create an account? Write to me directly.

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-calc2-antiderivatives. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: antiderivatives and the indefinite integral (201-SN3-RE)

The essentials

One antiderivative, one family, one constant

The only two rules, and the ones that do not exist

Composite forms, the ones you read at a glance

The rules in table form

The common antiderivatives

Composite forms: the u′u'u′ must be there

The mistakes that cost marks

1. Forgetting the constant of integration

2. Integrating a product factor by factor

3. Writing the antiderivative of one over x without an absolute value

4. Losing the dxdxdx while substituting

5. Forcing a substitution when the inner derivative is missing

6. Not returning to the original variable

7. Believing that every elementary function has an elementary antiderivative

8. Simplifying an antiderivative without looking at the domain

Which method to choose

Will a substitution work?

How the answer is expected to be written

A substitution, written up for the marker

Check before you hand in

The typical problem, taken apart

A substitution that resists, as on the first exam

Learn by heart

Frequently asked questions

Why do you have to add a constant to an antiderivative?

Is there a product rule for integrals?

How do you know if a substitution will work?

Why do we write an absolute value in the antiderivative of one over x?

Do all functions have an antiderivative you can compute?

Corrected exercises: Antiderivatives and the indefinite integral, 201-SN3-RE

See also

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Composite forms: the $u'$ must be there

4. Losing the $dx$ while substituting